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Difference between revisions of "number system"

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[[File:number systems.svg|right|300px]]
 
A '''number system''' is a mathematical notation for representing numbers of a given [[number set|set]]. They are the foundation for conveying, quantifying, and manipulating [[data]].
 
A '''number system''' is a mathematical notation for representing numbers of a given [[number set|set]]. They are the foundation for conveying, quantifying, and manipulating [[data]].
  
−
Number systems are mainly classified according to [[number notation|notations]] ([[positional notation]] vs [[sign-value notation]]) and their [[base]]. Today, we largely use the [[Arabic numerals]] which is a [[base-10]] positional notation numbering system. Machines on the other hand may use a different number system - such as the [[base-2]].
+
Number systems are mainly classified according to [[number notation|notations]] ([[positional notation]] vs [[sign-value notation]]) and their [[base]]. Today, we largely use the [[Arabic numerals]] which is a [[base-10]] positional notation numbering system. Machines on the other hand may use a different number system - such as the [[binary number system]].
  
 
== Notation ==
 
== Notation ==
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::<math>
 
::<math>
−
\begin{eqnarray*}
+
\begin{align}
−
N_{10} &=& 49835.5825 \\
+
N_{10} &= 49835.5825 \\
−
     &=& (d_4 d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3} d_{-4})_{10} \\
+
     &= (d_4 d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3} d_{-4})_{10} \\
−
     &=& 4 \times 10^4 + 9 \times 10^3 + 8 \times 10^2 + 3 \times 10^1 + 5 \times 10^0 + 5 \times 10^{-1} + 8 \times 10^{-2} + 2 \times 10^{-3} + 5 \times 10^{-4} \\
+
     &= 4 \times 10^4 + 9 \times 10^3 + 8 \times 10^2 + 3 \times 10^1 + 5 \times 10^0 + 5 \times 10^{-1} + 8 \times 10^{-2} + 2 \times 10^{-3} + 5 \times 10^{-4} \\
−
     &=& 40000 + 9000 + 800 + 30 + 5 + 0.5 + 0.08 + 0.002 + 0.0005 \\
+
     &= 40000 + 9000 + 800 + 30 + 5 + 0.5 + 0.08 + 0.002 + 0.0005 \\
−
     &=& 49835.5825_{10}
+
     &= 49835.5825_{10}
−
\end{eqnarray*}
+
\end{align}
 
</math>
 
</math>
  
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::<math>
 
::<math>
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\begin{eqnarray*}
+
\begin{align}
−
N_{10} &=& 0.938 \\
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N_{10} &= 0.938 \\
−
     &=& (\cdot d_{-1} d_{-2} d_{-3})_{10} \\
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     &= (d_{-1} d_{-2} d_{-3})_{10} \\
−
     &=& 9 \times 10^{-1} + 3 \times 10^{-2} + 8 \times 10^{-3} \\
+
     &= 9 \times 10^{-1} + 3 \times 10^{-2} + 8 \times 10^{-3} \\
−
     &=& 0.9 + 0.03 + 0.008 \\
+
     &= 0.9 + 0.03 + 0.008 \\
−
     &=& 0.938_{10}
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     &= 0.938_{10}
−
\end{eqnarray*}
+
\end{align}
 
</math>
 
</math>
  
−
=== Decimal Number System ===
+
=== Binary Number System ===
 
{{main|binary}}
 
{{main|binary}}
 
In the binary number system, the [[radix]] is 2 - i.e. a number system capable of only representing two discrete values: <math>\{0,1\}</math>. Let's consider the following number <math>10101110.011_{2}</math>. Note that the subscript is ''2''. We can express this [[binary number]] in polynomial form as follows:
 
In the binary number system, the [[radix]] is 2 - i.e. a number system capable of only representing two discrete values: <math>\{0,1\}</math>. Let's consider the following number <math>10101110.011_{2}</math>. Note that the subscript is ''2''. We can express this [[binary number]] in polynomial form as follows:
  
 
::<math>
 
::<math>
−
\begin{eqnarray*}
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\begin{align}
−
N_{2} &=& 10101110.011 \\
+
N_{2} &= 10101110.011 \\
−
     &=& (d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3})_{2} \\
+
     &= (d_7 d_6 d_5 d_4 d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3})_{2} \\
−
     &=& 1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3} \\
+
     &= 1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3} \\
−
     &=& 128 + 32 + 8 + 4 + 2 + 0.25 + 0.125 \\
+
     &= 128 + 32 + 8 + 4 + 2 + 0.25 + 0.125 \\
−
     &=& 174.375_{10} \\
+
     &= 174.375_{10} \\
−
\end{eqnarray*}
+
\end{align}
 
</math>
 
</math>
  
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::<math>
 
::<math>
−
\begin{eqnarray*}
+
\begin{align}
−
N_{8} &=& 703.104 \\
+
N_{8} &= 703.104 \\
−
     &=& (d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3})_{16} \\
+
     &= (d_2 d_1 d_0 \cdot d_{-1} d_{-2} d_{-3})_{16} \\
−
     &=& 7 \times 8^2 + 0 \times 8^1 + 3 \times 8^0 + 1 \times 8^{-1} + 0 \times 8^{-2} + 4 \times 8^{-3} \\
+
     &= 7 \times 8^2 + 0 \times 8^1 + 3 \times 8^0 + 1 \times 8^{-1} + 0 \times 8^{-2} + 4 \times 8^{-3} \\
−
     &=& 448 + 3 + 0.125 + 0.0078125 \\
+
     &= 448 + 3 + 0.125 + 0.0078125 \\
−
     &=& 451.1328125_{10}
+
     &= 451.1328125_{10}
−
\end{eqnarray*}
+
\end{align}
 
</math>
 
</math>
  
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::<math>
 
::<math>
−
\begin{eqnarray*}
+
\begin{align}
−
N_{16} &=& FDAE.F5F \\
+
N_{16} &= FDAE.F5F \\
−
     &=& (d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2})_{16} \\
+
     &= (d_3 d_2 d_1 d_0 \cdot d_{-1} d_{-2})_{16} \\
−
     &=& F \times 16^3 + D \times 16^2 + A \times 16^1 + E \times 16^0 + F \times 16^{-1} + 5 \times 16^{-2} + F \times 16^{-3} \\
+
     &= F \times 16^3 + D \times 16^2 + A \times 16^1 + E \times 16^0 + F \times 16^{-1} + 5 \times 16^{-2} + F \times 16^{-3} \\
−
     &=& 15 \times 16^3 + 13 \times 16^2 + 10 \times 16^1 + 14 \times 16^0 + 15 \times 16^{-1} + 5 \times 16^{-2} + 15 \times 16^{-3} \\
+
     &= 15 \times 16^3 + 13 \times 16^2 + 10 \times 16^1 + 14 \times 16^0 + 15 \times 16^{-1} + 5 \times 16^{-2} + 15 \times 16^{-3} \\
−
     &=& 61440  + 3328 + 160 + 15 + 0.9375 + 0.01953125 + 0.003662109375 \\
+
     &= 61440  + 3328 + 160 + 15 + 0.9375 + 0.01953125 + 0.003662109375 \\
−
     &\cong& 64943.9606934_{10}
+
     &\cong 64943.9606934_{10}
−
\end{eqnarray*}
+
\end{align}
 
</math>
 
</math>
  
−
Note that the final value is rounded as a result of fraction estimation. Note that just like [[octal]] is also a power of 2 (<math>2^4</math>), meaning conversion from binary to hex can be done by simply breaking down a [[binary number]] into groups of 4 bits and converting each group individually. For example, consider the number <math>101101011110001_2<math> which can be broken down into <math>[0101]_2[1010]_2[1111]_2[0001]_2 = [5]_{16}[A]_{16}[F]_{16}[1]_{16} = 5AF1_16<math>. Converting a number back from hex to binary is done by simply converting each individual hex digit back into a group of 4 bits.
+
Note that the final value is rounded as a result of fraction estimation. Note that just like [[octal]] is also a power of 2 (<math>2^4</math>), meaning conversion from binary to hex can be done by simply breaking down a [[binary number]] into groups of 4 bits and converting each group individually. For example, consider the number <math>101101011110001_2</math> which can be broken down into <math>[0101]_2[1010]_2[1111]_2[0001]_2 = [5]_{16}[A]_{16}[F]_{16}[1]_{16} = 5AF1_{16}</math>. Converting a number back from hex to binary is done by simply converting each individual hex digit back into a group of 4 bits.

Latest revision as of 06:08, 2 December 2015

number systems.svg

A number system is a mathematical notation for representing numbers of a given set. They are the foundation for conveying, quantifying, and manipulating data.

Number systems are mainly classified according to notations (positional notation vs sign-value notation) and their base. Today, we largely use the Arabic numerals which is a base-10 positional notation numbering system. Machines on the other hand may use a different number system - such as the binary number system.

Notation[edit]

In the positional notation, numbers are a string consisting of one or more juxtaposed digits. For example, fixed-point form number N takes the following form:

Equation upper N Subscript r Baseline equals left-parenthesis d Subscript n minus 2 Baseline d Subscript n minus 1 Baseline ellipsis d 2 d 1 d 0 dot d Subscript negative 1 Baseline d Subscript negative 2 Baseline d Subscript negative 3 Baseline ellipsis d Subscript negative m Baseline right-parenthesis Subscript r

Where r is the radix or base. The base represents the number of digits in the number system. d are the digits defined for that given radix. The dot in the center (⋅) is known as the radix point. It separates the integer part on the left from the fractional portion on the right. Furthermore, the value of number N in base r can be represented in polynomial form as

Equation upper N Subscript r Baseline equals sigma-summation Underscript i equals negative m Overscript n minus 1 Endscripts d Subscript i Baseline r Superscript i Baseline equals left-parenthesis d Subscript n minus 2 Baseline r Superscript n minus 2 Baseline plus d Subscript n minus 1 Baseline r Superscript n minus 1 Baseline plus ellipsis plus d 1 r Superscript 1 Baseline plus d 0 r Superscript 0 Baseline plus d Subscript negative 1 Baseline r Superscript negative 1 Baseline plus d Subscript negative 2 Baseline r Superscript negative 2 Baseline plus d Subscript negative m Baseline r Superscript negative m Baseline right-parenthesis Subscript r

Where d is the digit in the ith position. When a number has no fractional portion, the number can be more accurately called an integer. Conversely, if a number has no integer part, it is called a fractional number or simply a fraction.

On various occasions, a number might be written in the form Equation upper N Subscript r where the subscript r is the radix of number N. Subscripts are used to indicate a base; often, to avoid confusion, when presenting values in different basis.

Decimal Number System[edit]

Main article: decimal

For the most part, humans use the decimal number system (base-10) for all day-to-day activities. It is the number system we learn and grow up with hence the most familiar to us. In a base-10 number system there are 10 distinct digits: Equation StartSet 0 comma 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 EndSet . Consider the following number Equation 49835.5825 Subscript 10 . Note the 10. We can express that number in polynomial form as

Equation StartLayout 1st Row 1st Column upper N 10 2nd Column equals 49835.5825 2nd Row 1st Column Blank 2nd Column equals left-parenthesis d 4 d 3 d 2 d 1 d 0 dot d Subscript negative 1 Baseline d Subscript negative 2 Baseline d Subscript negative 3 Baseline d Subscript negative 4 Baseline right-parenthesis Subscript 10 Baseline 3rd Row 1st Column Blank 2nd Column equals 4 times 10 Superscript 4 Baseline plus 9 times 10 cubed plus 8 times 10 squared plus 3 times 10 Superscript 1 Baseline plus 5 times 10 Superscript 0 Baseline plus 5 times 10 Superscript negative 1 Baseline plus 8 times 10 Superscript negative 2 Baseline plus 2 times 10 Superscript negative 3 Baseline plus 5 times 10 Superscript negative 4 Baseline 4th Row 1st Column Blank 2nd Column equals 40000 plus 9000 plus 800 plus 30 plus 5 plus 0.5 plus 0.08 plus 0.002 plus 0.0005 5th Row 1st Column Blank 2nd Column equals 49835.5825 Subscript 10 EndLayout

Now consider the fractional decimal number Equation 0.938 Subscript 10 . Since there is no integer part, our exponents start at 0 and decrease with each additional digit:

Equation StartLayout 1st Row 1st Column upper N 10 2nd Column equals 0.938 2nd Row 1st Column Blank 2nd Column equals left-parenthesis d Subscript negative 1 Baseline d Subscript negative 2 Baseline d Subscript negative 3 Baseline right-parenthesis Subscript 10 Baseline 3rd Row 1st Column Blank 2nd Column equals 9 times 10 Superscript negative 1 Baseline plus 3 times 10 Superscript negative 2 Baseline plus 8 times 10 Superscript negative 3 Baseline 4th Row 1st Column Blank 2nd Column equals 0.9 plus 0.03 plus 0.008 5th Row 1st Column Blank 2nd Column equals 0.938 Subscript 10 EndLayout

Binary Number System[edit]

Main article: binary

In the binary number system, the radix is 2 - i.e. a number system capable of only representing two discrete values: Equation StartSet 0 comma 1 EndSet . Let's consider the following number Equation 10101110.011 Subscript 2 . Note that the subscript is 2. We can express this binary number in polynomial form as follows:

Equation StartLayout 1st Row 1st Column upper N 2 2nd Column equals 10101110.011 2nd Row 1st Column Blank 2nd Column equals left-parenthesis d 7 d 6 d 5 d 4 d 3 d 2 d 1 d 0 dot d Subscript negative 1 Baseline d Subscript negative 2 Baseline d Subscript negative 3 Baseline right-parenthesis Subscript 2 Baseline 3rd Row 1st Column Blank 2nd Column equals 1 times 2 Superscript 7 Baseline plus 0 times 2 Superscript 6 Baseline plus 1 times 2 Superscript 5 Baseline plus 0 times 2 Superscript 4 Baseline plus 1 times 2 cubed plus 1 times 2 squared plus 1 times 2 Superscript 1 Baseline plus 0 times 2 Superscript 0 Baseline plus 0 times 2 Superscript negative 1 Baseline plus 1 times 2 Superscript negative 2 Baseline plus 1 times 2 Superscript negative 3 Baseline 4th Row 1st Column Blank 2nd Column equals 128 plus 32 plus 8 plus 4 plus 2 plus 0.25 plus 0.125 5th Row 1st Column Blank 2nd Column equals 174.375 Subscript 10 EndLayout

Note that as part of the process we've actually converted the number to decimal.

Octal Number System[edit]

Main article: octal

The octal number system as the name implies uses radix of 8; i.e. the number system is represented via 8 discrete values: Equation StartSet 0 comma 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 EndSet . Let's take the number Equation 703.104 Subscript 8 ,

Equation StartLayout 1st Row 1st Column upper N 8 2nd Column equals 703.104 2nd Row 1st Column Blank 2nd Column equals left-parenthesis d 2 d 1 d 0 dot d Subscript negative 1 Baseline d Subscript negative 2 Baseline d Subscript negative 3 Baseline right-parenthesis Subscript 16 Baseline 3rd Row 1st Column Blank 2nd Column equals 7 times 8 squared plus 0 times 8 Superscript 1 Baseline plus 3 times 8 Superscript 0 Baseline plus 1 times 8 Superscript negative 1 Baseline plus 0 times 8 Superscript negative 2 Baseline plus 4 times 8 Superscript negative 3 Baseline 4th Row 1st Column Blank 2nd Column equals 448 plus 3 plus 0.125 plus 0.0078125 5th Row 1st Column Blank 2nd Column equals 451.1328125 Subscript 10 EndLayout

An interesting property of a base-8 system being Equation 2 cubed is that converting a number from binary to octal can easily done by breaking the number number into groups of 3 bits and converting each group individually into octal. Octal being 8 values means each group of 3 binary bits equates to a single octal digit. For example consider the following number Equation 1010011101 Subscript 2 . We can break the number into groups of 3 bits: Equation left-bracket 001 right-bracket Subscript 2 Baseline left-bracket 010 right-bracket Subscript 2 Baseline left-bracket 011 right-bracket Subscript 2 Baseline left-bracket 101 right-bracket Subscript 2 . Each of those groups can now be converted to octal independently: Equation left-bracket 1 right-bracket Subscript 8 Baseline left-bracket 2 right-bracket Subscript 8 Baseline left-bracket 3 right-bracket Subscript 8 Baseline left-bracket 5 right-bracket Subscript 8 Baseline equals 1235 Subscript 8 . Note that the conversion from octal back to binary can just as easily be done by simple converting each individual octal digit back into a group of 3 binary digits.

Hexadecimal Number System[edit]

Main article: Hexadecimal

The Hexadecimal number system or base-16 has 16 discrete digits. Those digits have traditionally been Equation StartSet 0 comma 1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 comma 9 comma upper A comma upper B comma upper C comma upper D comma upper E comma upper F EndSet . Note that A through F are decimal 10 through 15, respectively. In this example we'll use the hex number Equation upper F upper D upper A upper E period upper F Baseline 5 upper F 16 .

Equation StartLayout 1st Row 1st Column upper N 16 2nd Column equals upper F upper D upper A upper E period upper F Baseline 5 upper F 2nd Row 1st Column Blank 2nd Column equals left-parenthesis d 3 d 2 d 1 d 0 dot d Subscript negative 1 Baseline d Subscript negative 2 Baseline right-parenthesis Subscript 16 Baseline 3rd Row 1st Column Blank 2nd Column equals upper F times 16 cubed plus upper D times 16 squared plus upper A times 16 Superscript 1 Baseline plus upper E times 16 Superscript 0 Baseline plus upper F times 16 Superscript negative 1 Baseline plus 5 times 16 Superscript negative 2 Baseline plus upper F times 16 Superscript negative 3 Baseline 4th Row 1st Column Blank 2nd Column equals 15 times 16 cubed plus 13 times 16 squared plus 10 times 16 Superscript 1 Baseline plus 14 times 16 Superscript 0 Baseline plus 15 times 16 Superscript negative 1 Baseline plus 5 times 16 Superscript negative 2 Baseline plus 15 times 16 Superscript negative 3 Baseline 5th Row 1st Column Blank 2nd Column equals 61440 plus 3328 plus 160 plus 15 plus 0.9375 plus 0.01953125 plus 0.003662109375 6th Row 1st Column Blank 2nd Column approximately-equals 64943.9606934 Subscript 10 EndLayout

Note that the final value is rounded as a result of fraction estimation. Note that just like octal is also a power of 2 ( Equation 2 Superscript 4 ), meaning conversion from binary to hex can be done by simply breaking down a binary number into groups of 4 bits and converting each group individually. For example, consider the number Equation 101101011110001 Subscript 2 which can be broken down into Equation left-bracket 0101 right-bracket Subscript 2 Baseline left-bracket 1010 right-bracket Subscript 2 Baseline left-bracket 1111 right-bracket Subscript 2 Baseline left-bracket 0001 right-bracket Subscript 2 Baseline equals left-bracket 5 right-bracket Subscript 16 Baseline left-bracket upper A right-bracket Subscript 16 Baseline left-bracket upper F right-bracket Subscript 16 Baseline left-bracket 1 right-bracket Subscript 16 Baseline equals 5 upper A upper F 1 Subscript 16 . Converting a number back from hex to binary is done by simply converting each individual hex digit back into a group of 4 bits.

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